题目大意
给定一个(n)，求出点数为(n)的边双连通图的个数。
思路
其实思路跟点双连通分量计数差不多的。
我们设(F(x))为有标号无向图的指数级生成函数，(G(x))为有标号无向连通图的指数型生成函数。可以得到:
[F(x)=sum_{i=1}^{infty} frac{2^{inom{i}{2}}}{i!}x^i
]
[F(x)=e^{G(x)}
ightarrow G(x)=ln F(z)
]
接着我们设(D(x))为有根无向连通图的指数型生成函数，(B(x))为有根无向边双连通分量的指数型生成函数，我们可以得到：
[D(x)=sum_{i=1} frac{b_ie^{iD(x)}}{i!}x^i
]
性感证明就是我们根所在的边双联通分量大小如果为(i)，那么就相当于把连通图通过一条边挂在(i)个点上面，而不是跟点双联通计数一样选一个点为连通图的根，就是(e^{iD(x)})，而边双联通分量又有(b_i)种方法。
于是，从上面的式子我们可以推得：
[D(x)=B(xe^{D(x)})
]
我们如果设(F(x)=xe^{D(x)})，则(D(x)=B(F(x)))，两边同时做复合逆，可以得到(B(x)=D(F^{-1}(x)))。于是，这里我们就可以使用拓展拉格朗日反演了:
[[x^n]B(x)=[x^n]D(F^{-1}(x))
]
[=frac{1}{n}[x^{-1}]D^{'}(x)F(x)^{-n}
]
[=frac{1}{n}[x^{n-1}]D^{'}(x)(frac{x}{F(x)})^n
]
[=frac{1}{n}[x^{n-1}]D^{'}(x)(frac{x}{xe^{D(x)}})^n
]
[=frac{1}{n}[x^{n-1}]D^{'}(x)e^{-nD(x)}
]
于是，我们就可以在(Theta(nlog n))的时间复杂度内解决这个问题。但是我常数似乎很大。。。
( ext {Code})
#include <bits/stdc++.h>
using namespace std;
#define Int register int
#define mod 998244353
#define Gii 332748118
#define ll long long
#define MAXN 300005
#define Gi 3
int quick_pow (int a,int b){
int res = 1;for (;b;b >>= 1,a = 1ll * a * a % mod) if (b & 1) res = 1ll * res * a % mod;
return res;
}
int limit,l,r[MAXN];
void NTT (int *a,int type){
for (Int i = 0;i < limit;++ i) if (i < r[i]) swap (a[i],a[r[i]]);
for (Int mid = 1;mid < limit;mid <<= 1){
int Wn = quick_pow (type == 1 ? Gi : Gii,(mod - 1) / (mid << 1));
for (Int R = mid << 1,j = 0;j < limit;j += R){
for (Int k = 0,w = 1;k < mid;++ k,w = 1ll * w * Wn % mod){
int x = a[j + k],y = 1ll * w * a[j + k + mid] % mod;
a[j + k] = (x + y) % mod,a[j + k + mid] = (x + mod - y) % mod;
}
}
}
if (type == 1) return ;
int Inv = quick_pow (limit,mod - 2);
for (Int i = 0;i < limit;++ i) a[i] = 1ll * a[i] * Inv % mod;
}
int c[MAXN];
void Solve (int len,int *a,int *b){
if (len == 1) return b[0] = quick_pow (a[0],mod - 2),void ();
Solve ((len + 1) >> 1,a,b);
limit = 1,l = 0;
while (limit < (len << 1)) limit <<= 1,l ++;
for (Int i = 0;i < limit;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
for (Int i = 0;i < len;++ i) c[i] = a[i];
for (Int i = len;i < limit;++ i) c[i] = 0;
NTT (c,1);NTT (b,1);
for (Int i = 0;i < limit;++ i) b[i] = 1ll * b[i] * (2 + mod - 1ll * c[i] * b[i] % mod) % mod;
NTT (b,-1);
for (Int i = len;i < limit;++ i) b[i] = 0;
}
void deravitive (int *a,int n){
for (Int i = 1;i <= n;++ i) a[i - 1] = 1ll * a[i] * i % mod;
a[n] = 0;
}
void inter (int *a,int n){
for (Int i = n;i >= 1;-- i) a[i] = 1ll * a[i - 1] * quick_pow (i,mod - 2) % mod;
a[0] = 0;
}
int b[MAXN];
void Ln (int *a,int n){
memset (b,0,sizeof (b));
Solve (n,a,b);deravitive (a,n);
while (limit <= n) limit <<= 1,l ++;
for (Int i = 0;i < limit;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
NTT (a,1),NTT (b,1);
for (Int i = 0;i < limit;++ i) a[i] = 1ll * a[i] * b[i] % mod;
NTT (a,-1),inter (a,n);
for (Int i = n + 1;i < limit;++ i) a[i] = 0;
}
int F0[MAXN];
void Exp (int *a,int *B,int n)
{
if (n == 1) return B[0] = 1,void ();
Exp (a,B,(n + 1) >> 1);
for (Int i = 0;i < limit;++ i) F0[i] = B[i];
Ln (F0,n);
F0[0] = (a[0] + 1 + mod - F0[0]) % mod;
for (Int i = 1;i < n;++ i) F0[i] = (a[i] + mod - F0[i]) % mod;
NTT (F0,1);NTT (B,1);
for (Int i = 0;i < limit;++ i) B[i] = 1ll * F0[i] * B[i] % mod;
NTT (B,-1);
for (Int i = n;i < limit;++ i) B[i] = 0;
}
int read ()
{
int x = 0;char c = getchar();int f = 1;
while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}
while (c >= '0' && c <= '9'){x = (x << 3) + (x << 1) + c - '0';c = getchar();}
return x * f;
}
void write (int x)
{
if (x < 0){x = -x;putchar ('-');}
if (x > 9) write (x / 10);
putchar (x % 10 + '0');
}
int fac[MAXN],caf[MAXN],lim = 140000;
void init (){
fac[0] = 1;for (Int i = 1;i <= lim;++ i) fac[i] = 1ll * fac[i - 1] * i % mod;
caf[lim] = quick_pow (fac[lim],mod - 2);for (Int i = lim;i;-- i) caf[i - 1] = 1ll * caf[i] * i % mod;
}
int H[MAXN],H_[MAXN],G[MAXN],FG[MAXN],SG[MAXN];
void makerev (int len){
limit = 1,l = 0;
while (limit < len) limit <<= 1,l ++;
for (Int i = 0;i < limit;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << l - 1);
}
void prepare (){
int len = 1 << 17;makerev (len);
for (Int i = 0;i < len;++ i) H[i] = 1ll * quick_pow (2,1ll * i * (i - 1) / 2 % (mod - 1)) * caf[i] % mod;
Ln (H,len - 1);
for (Int i = 0;i < len;++ i) H[i] = H_[i] = 1ll * H[i] * i % mod;
deravitive (H_,len - 1),makerev (len << 1),NTT (H_,1);
}
void work (int n){
int len = 1 << 17;
memset (SG,0,sizeof (SG)),memset (F0,0,sizeof (F0));
for (Int i = 0;i < len;++ i) G[i] = 1ll * H[i] * (mod - n) % mod;
Exp (G,SG,len),makerev (len << 1),NTT (SG,1);
for (Int i = 0;i < len << 1;++ i) SG[i] = 1ll * SG[i] * H_[i] % mod;
NTT (SG,-1);
write (1ll * SG[n - 1] * quick_pow (n,mod - 2) % mod * fac[n - 1] % mod),putchar ('
');
}
signed main(){
init (),prepare ();
for (Int i = 1;i <= 5;++ i) work (read ());
return 0;
}